Lecture 23 Branch-and-Bound Algorithm. November 3, 2009

Transcription

1 Branch-and-Bound Algorithm November 3, 2009

2 Outline Lecture 23 Modeling aspect: Either-Or requirement Special ILPs: Totally unimodular matrices Branch-and-Bound Algorithm Underlying idea Terminology Formal description Algorithm demonstrated on an example Operations Research Methods 1

3 Either-Or Requirement Modeling Suppose you have two constrains and you can choose only one of them. For example, the constraints may represent two alternative resource consumption relations. Specifically suppose a company wants to introduce two new products 1 and 2, which can be made on two different machines. The machine production times are given by 3x 1 + 2x machine 1 x 1 + 4x machine 2 The company needs to purchase a machine and can afford to buy only one Operations Research Methods 2

4 machine. Thus, the requirement in the problem is either 3x 1 + 2x or x 1 + 4x How can we model this requirement within ILP setting? Operations Research Methods 3

5 One possibility: Lecture 23 Introduce a binary variable y {0, 1} for a decision of selecting machine: y = { 1 if we choose machine 1 0 otherwise Let M > 0 be very large number (e.g., M = 1000) We can model the either-or selection of machines as follows 3x 1 + 2x M(1 y) machine 1 x 1 + 4x My machine 2 A similar idea can be used for selection of k out of m constraints, where k and m are given positive integers with k < m Operations Research Methods 4

6 Solving ILPs Some of the ILPs have a special structure that allows us to ignore integer constraints and solve them as LPs and still obtain integer solutions Such a class of ILP problems is rather small; one such problem that we have seen is Transportation problem with integer supplies and demands Most of the ILP problems do not have the special structure These are solved using Branch-and-Bound (BB) algorithm Operations Research Methods 5

7 Transportation Problem: MG Auto model Denver Miami Supply Los Angeles Detroit New Orleans Demand minimize 80x x x x x x 33 subject to x 11 + x 12 = 1000 LA x 21 + x 22 = 1500 Detroit x 31 + x 32 = 1200 NewOrleans x 11 + x 21 + x 31 = 2300 Denver x 12 + x 22 + x 32 = 1400 Miami x ij 0 for all i, j Operations Research Methods 6

8 Let A be the matrix that captures the linearly independent constraints (the last row is linearly dependent on the first four rows) Ax = b represents the first four constraints What makes the transportation model special is the structure of the matrix A corresponding to the independent equality constraints Ax = b. A = The matrix is totally unimodular. DEF. A matrix A is totally unimodular if every square nonsingular submatrix has determinant +1 or -1. A square matrix is nonsingular if it is invertible. Operations Research Methods 7

9 A condition for an m n matrix A to be totally unimodular is: The matrix A is totally unimodular if it has the following properties (a) Every column of A contains at most two non-zero entries (b) Every entry in A is 0, +1, or -1 (c) The rows of A can be partitioned into two disjoint index sets B and C, such that (1) If two non-zero entries in a column of A have the same sign, then the row of one is in B, and the other is in C; (2) If two non-zero entries in a column of A have opposite signs, then the rows of both are in B, or both in C. A = Operations Research Methods 8

10 For our matrix A, we can see that the conditions are satisfied with B = {1, 2, 3} and C = {4}. Not all ILP problems are with such a nice stricture. To solve the general ILPs, we need a special algorithm that can deal with the integer nature of the variables. Operations Research Methods 9

11 Branch-and-Bound Algorithm Operations Research Methods 10

12 Suppose we are facing the problem maximize 9x 1 + 5x 2 + 6x 3 + 4x 4 subject to 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 1 + x 3 0 x 2 + x 4 0 x j {0, 1} j = 1,..., 4 There are 2 4 possibilities for the variables x j. We can enumerate them all and find the best decision. Operations Research Methods 11

13 One way to represent these options is through the use of a binary tree The places where the decision tree branches are referred as nodes. The nodes at the end of the branching tree are referred as leaf-nodes Our feasible points are at the leaf-nodes of this decision tree. We could build the whole tree and then explore the objective value at leaf-nodes But, this is not efficient. Branch-and-Bound (BB) algorithm is an efficient approach that builds the tree as needed and searches the nodes of the tree efficiently. Operations Research Methods 12

14 BB is an iterative algorithm that at each iteration branches the tree and possibly prunes the tree until the solution if found. Each node in a tree corresponds to a subproblem of the original problem. Each iteration involves the following operations: (a) Branching - deciding on from which node to branch among the active nodes in the tree (b) Bounding - operation that is performed at each created node to decide on which nodes to be fathomed (pruning the branches), It amounts to solving the corresponding subproblem at each node. (c) Fathoming - in essence pruning the tree. Based on some test and the bound values at the active nodes, the algorithm decides on nodes that can be discarded. This step also includes a termination criterion. Operations Research Methods 13

15 Initialization The initial node in the tree corresponds to solving the LP relaxation of the given problem LP relaxation is the problem resulting from the given ILP when we ignore integer constraints of the variables This node is active and it is the most recent. Operations Research Methods 14

16 Typical Iteration for MAXIMIZATION (a) Branching. Among the most recently created nodes that are active select one (braking ties according to the largest bound). Branch from that node by fixing one of the relaxed variables to 0 and 1 (in the node subproblem) (b) Bounding. At each new node, solve the corresponding LP problem and determine the optimal LP value. Round the non-integer value down (to the nearest integer). This is the bound of the subproblem at the node. (c) Fathoming. For each new node (subproblem) apply the following three tests: (1) Integer solution. If one of the new nodes has integer solution, its bound is compared to the bounds of other such nodes. If it does not have the best value - it is fathomed. If it has the best value it is fathomed and it is our current best solution (incumbent). (2) Bound value. If any of the new nodes has a bound smaller than currently the best bound - fathom the node. (3) Infeasibility. If LP at any of the new nodes has no solution (not feasible) - fathom the node. Operations Research Methods 15

17 Termination Stop when there are no more subproblems (no branching). The node with the best value provides the solution. Otherwise perform a new iteration. Operations Research Methods 16

18 Application BB algorithm to solve the following ILP: maximize 9x 1 + 5x 2 + 6x 3 + 4x 4 subject to 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 1 + x 3 0 x 2 + x 4 0 x j {0, 1} j = 1,..., 4 Operations Research Methods 17

19 Initialization At the initial node, we would solve its LP relaxation maximize 9x 1 + 5x 2 + 6x 3 + 4x 4 subject to 6x 1 + 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 1 + x 3 0 x 2 + x x j 1 j = 1,..., 4 Using LP algorithm, we find the optimal value (5/6, 1, 0, 1) and the optimal value z = If the solution and the optimal value were integers, we would stop. Since it is not, this value gives us 16 as the best lower bound on the optimal value of the ILP (blb = 16). Operations Research Methods 18

20 The initialization and moving to the first iteration Operations Research Methods 19

21 Iteration 1: We branch from the LP relaxation (ALL node) into two new nodes corresponding to x 1 = 1 and x 1 = 0. At node x 1 = 1 we solve the following LP maximize 9 + 5x 2 + 6x 3 + 4x 4 subject to 3x 2 + 5x 3 + 2x 4 4 x 3 + x 4 1 x 3 1 x 2 + x x j 1 j = 2,..., 4 Operations Research Methods 20

22 At node x 1 = 0 we solve the following LP maximize 5x 2 + 6x 3 + 4x 4 subject to 3x 2 + 5x 3 + 2x 4 10 x 3 + x 4 1 x 3 0 x 2 + x x j 1 j = 2,..., 4 Node to the right is incumbent and fathomed (no branching there ever). Node to the left is the only active node and most recent. Operations Research Methods 21

23 Iteration 2: We branch from the node x 1 = 1 going to nodes x 2 = 1 and x 2 = 0. At node x 2 = 1 (also we have x 1 = 1), we solve the following LP maximize x 3 + 4x 4 subject to 5x 3 + 2x 4 1 x 3 + x 4 1 x 3 1 x x j 1 j = 3, 4 The solution is (1, 1, 0, 1/2) and the optimal value is 16. Operations Research Methods 22

24 At node x 2 = 0 (and x 1 = 1) we solve the following LP maximize 9 + 6x 3 + 4x 4 subject to 5x 3 + 2x 4 4 x 3 + x 4 1 x 3 1 x x j 1 j = 3, 4 The solution is (1, 0, 4/5, 0) and the optimal value is We round this value down to 13. Operations Research Methods 23

25 The node to the right is fathomed since its value 13 is less than currently the best bound 16 Operations Research Methods 24